Solving Algebraic Equations: Finding \(x^5 + \frac{1}{x^5}\)

We are given the equation \(x^2 - \sqrt{7}x + 1 = 0\) and asked to find the value of \(x^5 + \frac{1}{x^5}\).

This type of problem often involves finding the value of \(x + \frac{1}{x}\) or \(x - \frac{1}{x}\) first and then using algebraic identities to find higher powers like \(x^2 + \frac{1}{x^2}\), \(x^3 + \frac{1}{x^3}\), and so on.

Step 1: Finding the value of \(x + \frac{1}{x}\)

Start with the given equation:

\(x^2 - \sqrt{7}x + 1 = 0\)

Since \(x=0\) is not a solution (as \(0^2 - \sqrt{7}(0) + 1 = 1 \neq 0\)), we can divide the entire equation by \(x\):

\(\frac{x^2}{x} - \frac{\sqrt{7}x}{x} + \frac{1}{x} = \frac{0}{x}\)

\(x - \sqrt{7} + \frac{1}{x} = 0\)

Rearranging the terms, we get:

\(x + \frac{1}{x} = \sqrt{7}\)

This is a crucial step as we can now use this value to find other related expressions.

Step 2: Calculating \(x^2 + \frac{1}{x^2}\)

We know the identity \((a+b)^2 = a^2 + b^2 + 2ab\). Let \(a = x\) and \(b = \frac{1}{x}\). Then \((x + \frac{1}{x})^2 = x^2 + (\frac{1}{x})^2 + 2(x)(\frac{1}{x})\).

Using the value from Step 1:

\((\sqrt{7})^2 = x^2 + \frac{1}{x^2} + 2(1)\)

\(7 = x^2 + \frac{1}{x^2} + 2\)

Subtract 2 from both sides:

\(x^2 + \frac{1}{x^2} = 7 - 2\)

\(x^2 + \frac{1}{x^2} = 5\)

Step 3: Calculating \(x^3 + \frac{1}{x^3}\)

We know the identity \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\). Let \(a = x\) and \(b = \frac{1}{x}\). Then \((x + \frac{1}{x})^3 = x^3 + (\frac{1}{x})^3 + 3(x)(\frac{1}{x})(x + \frac{1}{x})\).

Using the value from Step 1:

\((\sqrt{7})^3 = x^3 + \frac{1}{x^3} + 3(1)(x + \frac{1}{x})\)

\(7\sqrt{7} = x^3 + \frac{1}{x^3} + 3(\sqrt{7})\)

Subtract \(3\sqrt{7}\) from both sides:

\(x^3 + \frac{1}{x^3} = 7\sqrt{7} - 3\sqrt{7}\)

\(x^3 + \frac{1}{x^3} = 4\sqrt{7}\)

Step 4: Calculating \(x^5 + \frac{1}{x^5}\)

We can find \(x^5 + \frac{1}{x^5}\) using the values of \(x^2 + \frac{1}{x^2}\) and \(x^3 + \frac{1}{x^3}\). Consider the product \((x^2 + \frac{1}{x^2})(x^3 + \frac{1}{x^3})\):

\((x^2 + \frac{1}{x^2})(x^3 + \frac{1}{x^3}) = x^2 \cdot x^3 + x^2 \cdot \frac{1}{x^3} + \frac{1}{x^2} \cdot x^3 + \frac{1}{x^2} \cdot \frac{1}{x^3}\)

\(= x^{2+3} + x^{2-3} + x^{3-2} + \frac{1}{x^{2+3}}\)

\(= x^5 + x^{-1} + x^{1} + \frac{1}{x^5}\)

\(= x^5 + \frac{1}{x} + x + \frac{1}{x^5}\)

Rearranging the terms:

\((x^2 + \frac{1}{x^2})(x^3 + \frac{1}{x^3}) = (x^5 + \frac{1}{x^5}) + (x + \frac{1}{x})\)

Therefore, to find \(x^5 + \frac{1}{x^5}\), we can use the formula:

\(x^5 + \frac{1}{x^5} = (x^2 + \frac{1}{x^2})(x^3 + \frac{1}{x^3}) - (x + \frac{1}{x})\)

Substitute the values we found in Steps 1, 2, and 3:

• \(x + \frac{1}{x} = \sqrt{7}\)
• \(x^2 + \frac{1}{x^2} = 5\)
• \(x^3 + \frac{1}{x^3} = 4\sqrt{7}\)

\(x^5 + \frac{1}{x^5} = (5)(4\sqrt{7}) - (\sqrt{7})\)

\(x^5 + \frac{1}{x^5} = 20\sqrt{7} - \sqrt{7}\)

\(x^5 + \frac{1}{x^5} = (20 - 1)\sqrt{7}\)

\(x^5 + \frac{1}{x^5} = 19\sqrt{7}\)

Conclusion

The value of \(x^5 + \frac{1}{x^5}\) is \(19\sqrt{7}\).

Revision Table: Key Results

Additional Information: General Formulas for \(x^n + \frac{1}{x^n}\)

If \(x + \frac{1}{x} = k\), we can find higher powers using recursive formulas or direct identities:

• \(x^2 + \frac{1}{x^2} = k^2 - 2\)
• \(x^3 + \frac{1}{x^3} = k^3 - 3k\)
• \(x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2 = (k^2 - 2)^2 - 2\)
• \(x^5 + \frac{1}{x^5} = (x^2 + \frac{1}{x^2})(x^3 + \frac{1}{x^3}) - (x + \frac{1}{x}) = (k^2-2)(k^3-3k) - k\)

These formulas are very useful for solving problems involving powers of \(x + \frac{1}{x}\) or \(x - \frac{1}{x}\).

In this problem, \(k = \sqrt{7}\). Let's verify our results using these formulas:

• \(x^2 + \frac{1}{x^2} = (\sqrt{7})^2 - 2 = 7 - 2 = 5\) (Matches Step 2)
• \(x^3 + \frac{1}{x^3} = (\sqrt{7})^3 - 3(\sqrt{7}) = 7\sqrt{7} - 3\sqrt{7} = 4\sqrt{7}\) (Matches Step 3)
• \(x^5 + \frac{1}{x^5} = (k^2-2)(k^3-3k) - k = ((\sqrt{7})^2-2)((\sqrt{7})^3-3\sqrt{7}) - \sqrt{7}\)
• \(= (7-2)(7\sqrt{7}-3\sqrt{7}) - \sqrt{7}\)
• \(= (5)(4\sqrt{7}) - \sqrt{7}\)
• \(= 20\sqrt{7} - \sqrt{7}\)
• \(= 19\sqrt{7}\) (Matches Step 4)

The results are consistent.